When does Newton's Second Law not hold?

  1. I know a high school student who has this question. My first thought was that it does not hold if the mass is changing with time. I was thinking of the situation of a snowball rolling down a mountain picking up more snow as it rolls to the bottom. There is obviously mass gained so does Newton's Second Law hold for this situation at the very basic level (for a high school physics question)? If it does, can you think of a situation where it wouldn't hold? Remember that this is a high school introductory course in Physics
     
  2. jcsd
  3. A.T.

    A.T. 6,286
    Science Advisor
    Gold Member

    Yes it does. You just have to use the instantaneous mass.

    Maybe in non-inertial reference frames, unless you introduce inertial forces to make it work again.
     
  4. Nathanael

    Nathanael 1,445
    Homework Helper

    http://en.wikipedia.org/wiki/Newton's_laws_of_motion#Newton.27s_second_law

    I think A.T. is right about using instantaneous mass, but a highschool student ("at the very basic level") would probably just learn what the quote from wikipedia says.
     
  5. HallsofIvy

    HallsofIvy 40,967
    Staff Emeritus
    Science Advisor

    If mass is changing rather than [itex] F= ma[/itex] use [itex]F= d(mv)/dt= m (dv/dt)+ (dm/dt)v= ma+ (dm/dt)v [itex].
     
    Last edited by a moderator: Jul 25, 2014
  6. According to wikipedia when mass is changing the equation that holds in the case of mass variable systems is [tex]F=m\frac{dv}{dt}-v_{rel}\frac{dm}{dt}[/tex]. This is different if we use instanteneous mass only and is not exactly the same as dp/dt because [itex]v_{rel}[/itex] is the relative velocity of the incoming or outgoing mass relative to the c.o.m of the main body.

    http://en.wikipedia.org/wiki/Variable-mass_system
     
  7. sophiecentaur

    sophiecentaur 14,186
    Science Advisor
    Gold Member

    A plague on all smartarse high school questions that are given to students without a thought as to their reaction. People who ask unanswerable questions of their students should be locked in a room with a dozen such questions (aimed just above their level) and not let out till they have answered them. (No access to PF either!!)
     
  8. Well maybe you are right hehe, at least those type of questions shouldnt put in written tests. It is just an extra term added though in Newton's Law familiar form F=ma.
     
  9. A.T.

    A.T. 6,286
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    Gold Member

    Yes, right. The force depends on how much you have to accelerate the gained mass, to be at the same speed as the original mass.
     
  10. Jano L.

    Jano L. 1,274
    Gold Member

    This is quite widespread but erroneous idea, perhaps originating in the custom of writing Newton's second law in the form
    $$
    F=\frac{dp}{dt}.
    $$
    But this form is valid only for systems of constant mass. If the body considered loses parts so its mass is changing, the proper equation of motion for the center of mass of such changing body is
    $$
    F_{ext} + F_{exh} =m\frac{dv}{dt}
    $$
    where ##F_{ext}## is the usual external force (e.g. gravity) and ##F_{exh}## is the reaction force on the body due to the parts leaving it (e.g. exhaust gas). This equation has the form
    $$
    F=ma,
    $$
    not
    $$F=dp/dt.
    $$
    In this sense, the formula ##F=ma## is actually more general than ##F = dp/dt##. We should therefore teach Newton's second law to students by always writing it in the form ##F=ma##. Not only is it more informative, but by appropriate choice of expression for the total force ##F##, applicable even to systems of variable mass.
     
  11. sophiecentaur

    sophiecentaur 14,186
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    Gold Member

    The question would be quite a valid one if delivered verbally / interactively and more in the form of "Can you think of situations where N2 is not quite adequate for solving a problem". This could be followed by appropriate prompting and could provide an enlightening and fun session.

    PS I wish contributors would make more of an effort to address questions on the level they were originally posed. This thread was about Introductory High School Science - in which Calculus is probably quite a stretch for most students. There is plenty of opportunity to show us what you know, elsewhere (big boys' threads). It makes things much more approachable for 'beginners' when they aren't swamped by high level content. I know I can also be guilty of this; it's a big temptation.
     
    1 person likes this.
  12. Jano, I think you're citing the same formula as Delta [tex]F=m\frac{dv}{dt}-v_{rel}\frac{dm}{dt}[/tex] except you've relabeled ##v_{rel}\frac{dm}{dt}## as ##F_{exh}##. I'm not sure if it is right to call ##v_{rel}\frac{dm}{dt}## a force. (I sincerely don't know.) If it's an external force of the lost/gained mass acting on the body's center of mass, then wouldn't you also include it in the term ##F_{ext}## when adding up the individual forces to obtain the net external force? If not, why not? Perhaps it's not an external force because there's only one m variable in the entire equation? Any guidance is appreciated.

    Here's the article Wikipedia cites (worth reading the first half of the 2nd page): http://articles.adsabs.harvard.edu/full/seri/CeMDA/0053//0000227.000.html
     
    Last edited: Jul 24, 2014
  13. Jano L.

    Jano L. 1,274
    Gold Member

    I meant a more general description, where direction of the leaving parts is not specified. Then the force due to departing parts may not be expressible in such a simple manner. The formula you refer to is applicable only to special case where the departing parts move in a direction opposite to the velocity of the body.

    The equation of motion is meant for the motion of the center of mass of the remaining body. This remaining body is acted upon by two agents - the ordinary external force such as gravity or say, a pulling rope, and the force due to parts that have left the body. These parts are to be regarded as extraneous to the body so that no cancellation of force pairs occurs and non-zero net force due to the leaving parts is accounted for. This force is present only once in the equation of motion, of course. It would be a mistake to count it twice by including it to ##F_{ext}## and to ##F_{exh}##. If you wish, you may use different concept of external force, where ##F_{ext}## includes the force due to leaving parts and discard ##F_{exh}##.
     
  14. +1 to that
     
  15. Is this just your personal opinion or can you proof it?
     
  16. stevendaryl

    stevendaryl 3,343
    Science Advisor

    I seem to remember a long thread about this topic a few months back.
     
  17. Jano L.

    Jano L. 1,274
    Gold Member

    For systems of constant mass, the equation
    $$
    \mathbf F_{ext}=\frac{d\mathbf p}{dt}
    $$
    with ##\mathbf p =m\mathbf v## is generally valid because it is equivalent to ##\mathbf F_{ext}=m\mathbf a##, which we know is valid for such systems.

    For systems of variable mass, the equation

    $$
    \mathbf F_{ext}=\frac{d\mathbf p}{dt}
    $$
    with ##\mathbf p =m\mathbf v## is generally invalid. Why? First, it does not take into account direction in which the leaving parts move. But this is important for the resulting motion. Also, this equation can be rewritten into form
    $$
    \mathbf F_{ext} = \frac{dm}{dt}\mathbf v + m\frac{d\mathbf v }{dt}
    $$
    which shows it is inconsistent with the Galilei relativity principle: the term ## \frac{dm}{dt}\mathbf v## depends on the frame of reference but the other terms do not. Furthermore, this equation leads to motion inconsistent with the law of conservation of momentum of isolated system. This law leads to the correct equation of motion
    $$
    \mathbf F_{ext} + \mathbf F_{exh} = m\frac{d\mathbf v}{dt}
    $$
    where ##\mathbf F_{ext}## is the force due to external bodies and ##\mathbf F_{exh} ## is the force due to leaving parts (exhaust gases in the case of a rocket).
     
  18. stevendaryl

    stevendaryl 3,343
    Science Advisor

    I can't find it, now, but there was a long discussion about Newtonian physics as applied to variable-mass systems a few months ago. I think it's a very complicated subject, maybe too complicated to have a useful "law of motion".

    Here's a scenario where I don't think your formula gives the correct answer, either.

    Imagine you have an extended object, maybe a stick, or maybe a cigarette. With time, it's getting shorter, because one end is breaking off (or burning, or whatever). For simplicity, let the object be oriented in the y-direction, and assume that there are NO significant forces in the y-direction. There are no external forces at work, and the pieces that are breaking off or burning up just drop straight down (in the z-direction) without imparting any momentum to the remaining matter in the object.

    So any equation of motion of the form F = ma would presumably tell you that there is no acceleration in the y-direction. But that might not be true. If by "the location" of the object, you mean the location of the center of mass, then the location is changing with time, because when pieces break off, the center of mass of the remaining matter shifts. The equation of motion governing this shift of center of mass would depend on the details of how the crumbling or burning is propagating, and would have nothing much to do with forces in the y-direction.
     
  19. Nugatory

    Staff: Mentor

    That's a problem with the extended object; the momentum law that Jano L cited applies to a point mass emitting another point mass. Within your extended object, any infinitesimal element will obey Newton's second law.
     
  20. Apparently there is a catch with stevendaryls example, that equation holds for extended objects like a rocket where c.o.m of rocket changes because they lose fuel mass due to their burning to produce exhaust gases.
     
  21. Jano L.

    Jano L. 1,274
    Gold Member

    Interesting example. I did not think about such case, but I think now even here the equation applies. It is all about the meaning of ##\mathbf v##. The derivation of the equation introduces the quantity ##\mathbf v## as

    $$
    \mathbf v = \frac{\mathbf p}{m}
    $$

    where ##\mathbf p## is momentum of the remaining body and ##m## is mass of the remaining body. In your example with the cigarette, this ##\mathbf v## is constant so both sides of the equation of motion vanish.

    I did said above that ##\mathbf v## in the equation is velocity of the center of mass of the remaining body. Your example shows that sometimes this meaning is not exactly the same as the one given by the definition above.

    Luckily in rocket science, the difference is negligible, so we can use center of mass velocity in the equation without making too great an error :-)
     
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